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11TH · MATH

Question 2

menu_bookSolutionmenu_bookTheorymenu_book•(i)f(x)=x2−4xf(x)=x^2-4xf(x)=x2−4xmenu_book•(ii)f(x)=x2−5x+6f(x)=x^2-5x+6f(x)=x2−5x+6menu_book•(iii)f(x)=−x2+2x−8f(x)=-x^2+2x-8f(x)=−x2+2x−8menu_book•(iv)f(x)=x2−4x+4f(x)=x^2-4x+4f(x)=x2−4x+4menu_book•(v)f(x)=x2+2x−8.3f(x)=x^2+2x-8.3f(x)=x2+2x−8.3menu_book•(vi)f(x)=6−x−x2f(x)=6-x-x^2f(x)=6−x−x2
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11TH · MATH

Question 2

menu_bookSolutionmenu_bookTheorymenu_book•(i)f(x)=x2−4xf(x)=x^2-4xf(x)=x2−4xmenu_book•(ii)f(x)=x2−5x+6f(x)=x^2-5x+6f(x)=x2−5x+6menu_book•(iii)f(x)=−x2+2x−8f(x)=-x^2+2x-8f(x)=−x2+2x−8menu_book•(iv)f(x)=x2−4x+4f(x)=x^2-4x+4f(x)=x2−4x+4menu_book•(v)f(x)=x2+2x−8.3f(x)=x^2+2x-8.3f(x)=x2+2x−8.3menu_book•(vi)f(x)=6−x−x2f(x)=6-x-x^2f(x)=6−x−x2
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Accedmychevron_right11thchevron_rightmathchevron_rightTheory Of Quadratic Functionschevron_rightExercise 3.1

(ii)

Minimum point (52,−14)\boxed{\text{Minimum point }\left(\frac{5}{2},-\frac{1}{4}\right)}Minimum point (25​,−41​)​ Domain R,  Range [−14,∞)\boxed{\text{Domain }\mathbb{R},\;\text{Range }\left[-\frac{1}{4},\infty\right)}Domain R,Range [−41​,∞)​
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